# A curious relationship between lunar periods and the solar year

Earlier today I was exploring the astronomy features on WolframAlpha, when I stumbled upon an intriguing numerical relationship involving the values for the lunar synodic and sidereal months. Using the values

$$P_{syn}=1\,\text{synodic month}=29.530588\,\text{days},~~P_{sid}=1\,\text{sidereal month}=27.321661\,\text{days},$$

I noticed that this algebraic combination is very close to one year:

$$\frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}=365.256396\,\text{days}\\ \frac{1}{P_{sid}^{-1}-P_{syn}^{-1}}-1\,\text{tropical year}=0.0142053\,\text{days}\approx20\,\text{minutes}.$$

I really doubt this this is a coincidence. Is there a simple way to understand what makes this approximate equality hold?

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The sidereal year differs from the tropical year caused by the precession of Earth's rotation axis: en.wikipedia.org/wiki/Sidereal_year – Gerald Mar 24 '14 at 22:12

The synodic month is the "average period of the Moon's revolution with respect to the line joining the Sun and Earth". However, the Earth also moves in its orbit around the Sun during this month. From our vantage point, the Sun has appeared to move in the sky with respect to the background stars, in the same direction as the Moon moves in the sky with respect to the background stars.

Your calculation deals with the sidereal month, so your result is the calculation of the sidereal year.

When the Sun returns to the same spot in the sky, that is the sidereal year, whose length is 365.256363004 days, very close to your calculation. So why is that off from the tropical year by 20 minutes? Because the tropical year (the year that keeps the seasons in place throughout the calendar year) is slightly shorter than the sidereal year.

The tropical year is about 20 minutes shorter than the time it takes Earth to complete one full orbit around the Sun as measured with respect to the fixed stars (the sidereal year).

You just found the difference between the sidereal year and the tropical year, and that was no coincidence.

The number of sidereal months in a sidereal year is one more than the number of synodic months in a sidereal year. That is because the Earth goes around the Sun once of year (of course), leading to one less synodic month than sidereal month.

Here, $P_{syn}$ is the synodic period and $P_{sid}$ is the sidereal period, and $Y_{sid}$ is the sidereal year, all in days.

$$\frac{Y_{sid}}{P_{sid}} = \frac{Y_{sid}}{P_{syn}} + 1$$

Dividing both sides by $Y_{sid}$ yields:

$$\frac{1}{P_{sid}} = \frac{1}{P_{syn}} + \frac{1}{Y_{sid}}$$

Solving for $Y_{sid}$...

$$\frac{1}{P_{sid}} - \frac{1}{P_{syn}} = \frac{1}{Y_{sid}}$$

or

$$P_{sid}^{-1} - P_{syn}^{-1} = \frac{1}{Y_{sid}}$$

Multiplying both sides by $Y_{sid}$ and dividing both sides by $P_{sid}^{-1} - P_{syn}^{-1}$ yields

$$Y_{sid} = \frac{1}{P_{sid}^{-1} - P_{syn}^{-1}}$$

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I do not manage to understand how nor why a relation between sidereal and synodic MONTHS gives up the duration of the sidereal YEAR. – Envite Mar 25 '14 at 9:58
@Envite I have added a derivation of the formula for the sidereal year based on the relationship between the sidereal year, the sidereal period, and the synodic period. – rgettman Mar 25 '14 at 17:47
Now it makes all the sense :) Thanks – Envite Mar 25 '14 at 18:44
It took a while to sink in but this makes much more sense now that I know I should be comparing it to the sidereal year. Thanks. – David H Mar 26 '14 at 1:44